   Chapter 2.5, Problem 27E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its domain. State the domain.Theorem 4  Theorem 5 Theorem 7 Theorem 9  Q ( x ) = x − 2 3 x 3 − 2

To determine

To state: The domain and explain the function is continuous at every number in its domain.

Explanation

Given:

The function Q(x)=x23x32.

Theorems used:

4. If f and g are continuous at a and c is a constant, then the following functions are also continuous at a:

(i) f+g (ii) fg (iii) cf (iv) fg (v) fg if g(a)0.

5. (a) Any polynomial is continuous everywhere; that is, it is continuous on =(,).

(b) Any rational function is continuous wherever it is defined; that is, it is continuous on its domain.

7. The functions such as “Polynomials, rational functions, root functions, trigonometric functions, inverse trigonometric functions, exponential functions and logarithmic functions” are continuous at every number in their domains.

9. If g is continuous at a and f is continuous at g(a), then the composite function fg given by (fg)(x)=f(g(x)) is continuous at a.

Calculation:

The domain is the set of all input values of the function for which the function is real and defined.

Consider the denominator of the function Q(x) and equate to zero to obtain the undefined points.

Since the denominator of Q(x) is x32, the undefined points are obtained as shown below:

x32=0x3=2x=23x=23,23(1+3i2),23(13i2)

Thus, the real solution is, x=23.

Hence, the function is undefined when x=23.

Therefore, the domain of the function is, D={x|x23}.

And, the interval notation of the domain of Q(x) is (,23)(23,)

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